How to find expectation of independent white noise?

Question

$x_t=w_tw_{t-1}$, where $w_t$~$N(0,\sigma^2)$, $w_t$ uncorrelated.
$y_t=x^{2}_{t}$

$$E[y_t]=(\sigma^2)^2$$

Compute autocovariance $\gamma(h)$ of $y_t$, when $h=1$.
$$cov(y_t, y_{t+1})=cov(x^{2}_{t}, x^{2}_{t-1})$$ $$=cov(w_tw_{t-1}, w_{t+1}w_t)$$ $$=E[w_tw_{t-1}w_{t+1}w_t]-E[w_t]E[w_{t-1}]E[w_{t+1}]E[w_t]$$

This is the first term of the above,
$E[w_tw_{t-1}w_{t+1}w_t]=E[w_t]E[w_{t-1}]E[w_{t+1}]E[w_t]$

Is this wrong? why? $w_t$ are uncorrelated, I treat them as independent.

Answer 1

I figured it out. Because $w_t$ is correlated with itself, you can't break first $w_t$ and second $w_t$ into two separate expected values.

For instance, $$E[w_tw_t]=E[w^{2}_{t}]$$

Therefore, $$E[w_tw_{t-1}w_tw_{t-1}]=E[w^{2}_{t}]E[w^{2}_{t-1}]$$

Answer 2

For every $t$, $E(y_t^2)=E(w_{t-1}^4)E(w_t^4)=9\sigma^8$ and $E(y_t)=E(w_{t-1}^2)E(w_t^2)=\sigma^4$ hence $\gamma(0)=\mathrm{cov}(y_t,y_{t})=8\sigma^8$.

For every $t$, $E(y_ty_{t+1})=E(w_{t-1}^2)E(w_t^4)E(w_{t+1}^2)=3\sigma^8$ hence $\gamma(1)=\mathrm{cov}(y_t,y_{t+1})=2\sigma^8$.

For every $t$ and every $h\geqslant2$, $y_t$ and $y_{t+h}$ are independent hence $\gamma(h)=\mathrm{cov}(y_t,y_{t+h})=0$.